Страница 301 номер 1000, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Найти площадь криволинейной трапеции, ограниченной прямыми \(x = a\), \(x = b\), осью \(Ox\) и графиком функции \(y = f(x)\): 1) \(a = 2\), \(b = 4\), \(f(x) = x^3\); 2) \(a = 3\), \(b = 4\), \(f(x) = x^2\); 3) \(a = -2\), \(b = 1\), \(f(x) = x^2 + 1\); 4) \(a = 0\), \(b = 2\), \(f(x) = x^3 + 1\); 5) \(a = \frac{\pi}{3}\), \(b = \frac{2\pi}{3}\), \(f(x) = \sin x\); 6) \(a = -\frac{\pi}{6}\), \(b = 0\), \(f(x) = \cos x\).
\(S = \int_{a}^{b}{f(x)dx} = F(b) - F(a)\)
1) \(a = 2\), \(b = 4\), \(f(x) = x^{3}\)
\(S = \int_{2}^{4}{x^{3}dx} = \left. \ \dfrac{x^{4}}{4} \right|_{2}^{4} = \dfrac{4^{4}}{4} - \dfrac{2^{4}}{4} = 64 - 4 = 60;\)
2) \(a = 3\), \(b = 4\), \(f(x) = x^{2}\)
\(S = \int_{3}^{4}{x^{2}dx} = \left. \ \dfrac{x^{3}}{3} \right|_{3}^{4} = \dfrac{64}{3} - \dfrac{27}{3} = \dfrac{37}{3} = 12\dfrac{1}{3};\)
3) \(a = - 2\), \(b = 1\), \(f(x) = x^{2} + 1\)
\(S = \int_{-2}^{1}{\left( x^{2} + 1 \right)dx} = \left. \ \dfrac{x^{3}}{3} + x \right|_{-2}^{1} = \dfrac{1}{3} + 1 + \dfrac{8}{3} + 2 = 6;\)
4) \(a = 0\), \(b = 2\), \(f(x) = x^{3} + 1\)
\(S = \int_{0}^{2}{\left( x^{3} + 1 \right)dx} = \left. \ \dfrac{x^{4}}{4} + x \right|_{0}^{2} = 4 + 2 = 6;\)
5) \(a = \frac{\pi}{3}\), \(b = \frac{2\pi}{3}\), \(f(x) = \sin x\)
\(S = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}}{\sin xdx} = \left. \ - \cos x \right|_{\frac{\pi}{3}}^{\frac{2\pi}{3}} = \dfrac{1}{2} + \dfrac{1}{2} = 1;\)
6) \(a = - \frac{\pi}{6}\), \(b = 0\), \(f(x) = \cos x\)
\(S = \int_{-\frac{\pi}{6}}^{0}{\cos xdx} = \left. \ \sin x \right|_{-\frac{\pi}{6}}^{0} = 0 - \left(-\dfrac{1}{2} \right) = \dfrac{1}{2}.\)
